Let $\mathbf{a} = \begin{pmatrix} 7 \\ -4 \\ -4 \end{pmatrix}$ and $\mathbf{c} = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}.$  Find the vector $\mathbf{b}$ such that $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ are collinear, and $\mathbf{b}$ bisects the angle between $\mathbf{a}$ and $\mathbf{c}.$

[asy]
unitsize(0.5 cm);

pair A, B, C, O;

A = (-2,5);
B = (1,3);
O = (0,0);
C = extension(O, reflect(O,B)*(A), A, B);

draw(O--A,Arrow(6));
draw(O--B,Arrow(6));
draw(O--C,Arrow(6));
draw(interp(A,C,-0.1)--interp(A,C,1.1),dashed);

label("$\mathbf{a}$", A, NE);
label("$\mathbf{b}$", B, NE);
label("$\mathbf{c}$", C, NE);
[/asy]
Explanation: The line through $\mathbf{a}$ and $\mathbf{c}$ can be parameterized by
\[\begin{pmatrix} 7 - 9t \\ -4 + 3t \\ -4 + 6t \end{pmatrix}.\]Then $\mathbf{b}$ is of this form.  Furthermore, the angle between $\mathbf{a}$ and $\mathbf{b}$ is equal to the angle between $\mathbf{b}$ and $\mathbf{c}.$  Hence,
\[\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} = \frac{\mathbf{b} \cdot \mathbf{c}}{\|\mathbf{b}\| \|\mathbf{c}\|}.\]We can cancel the factors of $\|\mathbf{b}\|,$ to get
\[\frac{\begin{pmatrix} 7 \\ -4 \\ -4 \end{pmatrix} \cdot \begin{pmatrix} 7 - 9t \\ -4 + 3t \\ -4 + 6t \end{pmatrix}}{\left\| \begin{pmatrix} 7 \\ -4 \\ -4 \end{pmatrix} \right\|} = \frac{\begin{pmatrix} 7 - 9t \\ -4 + 3t \\ -4 + 6t \end{pmatrix} \cdot \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}}{\left\| \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} \right\|}.\]Then
\[\frac{(7)(7 - 9t) + (-4)(-4 + 3t) + (-4)(-4 + 6t)}{9} = \frac{(7 - 9t)(-2) + (-4 + 3t)(-1) + (-4 + 6t)(2)}{3}\]Solving, we find $t = \frac{3}{4}.$  Therefore, $\mathbf{b} = \boxed{\begin{pmatrix} 1/4 \\ -7/4 \\ 1/2 \end{pmatrix}}.$